This result is a consequence of some musings in the last week. A few basic results will need to be proven first. The main result is:
Example. The above result guarantees for any x∊ ℤ a solution such that y ∊ ℤ in the following equations:
Or more generally we have the following corollary for polynomials with integer terms:
I scribbled this Proposition (below) last week and wondered if it’s true in general. The proof turned out rather simple. I like this fact because of its general nature.
Proposition 1 For n∊ℤ+, k,x∊ℤ, and m∊ℕ.
The proof proceeds by showing first that given some integer of some parity, its powers remain of the same parity. For even, it follows immediately, for odd a simple argument using the Binomial Theorem shows that the remaining constant in an arbitrary expansion of 2k + 1 is 1, and the remining terms factorise as multiples of 2 i.e. (2k + 1)n = 2m +1 m∊ℤ.
The result from the top of the page is a direct corollary of Proposition 1.
Example. The above result guarantees for any x∊ ℤ a solution such that y ∊ ℤ in the following equations:
y = ½ ( ax3 + bx12 + cx81 + dx739 + ex993 - (a + b + c + d + e)x )
y = ½ ( a1x1 + a2x2 + a3x3 + ... + anx179 - (Σai)x )
Or more generally we have the following corollary for polynomials with integer terms:
I scribbled this Proposition (below) last week and wondered if it’s true in general. The proof turned out rather simple. I like this fact because of its general nature.
Proposition 1 For n∊ℤ+, k,x∊ℤ, and m∊ℕ.
xn + xn+m = 2k
The proof proceeds by showing first that given some integer of some parity, its powers remain of the same parity. For even, it follows immediately, for odd a simple argument using the Binomial Theorem shows that the remaining constant in an arbitrary expansion of 2k + 1 is 1, and the remining terms factorise as multiples of 2 i.e. (2k + 1)n = 2m +1 m∊ℤ.
The result from the top of the page is a direct corollary of Proposition 1.
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